Trig and Integrals

Hi there, this is my first post here and I'm hoping someone can help me out, I'm working on an assignment with integrals and while I can manage workng out the number versions of the question just fine I've now encountered Trig functions in integrals and I've hit a brick wall. Can anyone possibly help me get this stuff started? I need to know how to do this for my exam weeks next week. Thanks for any and all help.

Also remember that the integral of secxtanx is secx and that the integral of sec^2x is tanx.

There are actually methods which are helpful in solving trigonometic identities (at least, I was taught them when I was taking the calculus series), I can't remember them all right now. It should be in your calc book.

Actually, my approach to almost any problem involving trig functions is to rewrite in terms of sine and cosine only!
[tex]tan^5(x) Sec^3(X)= \frac{sin^5(x)}{cos^5(x)}\frac{1}{cos^5(x)}= \frac{sin^5(x)}{cos^8(x)}[/tex]
Since that involves an odd power of sin(x), Rewrite the integral as
[tex]\int\tan^5(x)sec^3(x)dx= \int\frac{sin^4(x)}{cos^8(x)}sin(x)dx=\int\frac{(1-cos^2(x))^4}{cos^8(x)}sin(x)dx[tex]
Now, let u= cos(x) so that du= -sin(x)dx and the integral becomes
[tex]\int\frac{(u^2- 1)^4}{u^8}du[/tex]

Or as island-boy has pointed out. When you need to tackle an integral with the product of odd power of tan, and a sec (the power of sec can be even or odd) function you can use the fact that:
[tex]\frac{d}{dx} \sec x = \tan x \sec x[/tex]
And the Trig Identity: tan2x + 1 = sec2x Or tan2x = sec2x - 1
It goes like this:
[tex]\tan ^ 5 x \sec ^3 x dx = \int (\tan ^ 4 x \sec ^ 2 x) (\tan x \sec x) dx[/tex]
Now make the substitution: u = sec x, and see if you can finish the problem. :)